python - Pandas DataFrame 窗口函数

Question

我正在尝试以类似于您使用 SQL 窗口函数的方式来操作我的数据框。考虑以下样本集:

import pandas as pd

df = pd.DataFrame({'fruit' : ['apple', 'apple', 'apple', 'orange', 'orange', 'orange', 'grape', 'grape', 'grape'],
               'test' : [1, 2, 1, 1, 2, 1, 1, 2, 1],
               'analysis' : ['full', 'full', 'partial', 'full', 'full', 'partial', 'full', 'full', 'partial'],
               'first_pass' : [12.1, 7.1, 14.3, 19.1, 17.1, 23.4, 23.1, 17.2, 19.1],
               'second_pass' : [20.1, 12.0, 13.1, 20.1, 18.5, 22.7, 14.1, 17.1, 19.4],
               'units' : ['g', 'g', 'g', 'g', 'g', 'g', 'g', 'g', 'g'],
               'order' : [2, 1, 3, 2, 1, 3, 3, 2, 1]})

+--------+------+----------+------------+-------------+-------+-------+| fruit  | test | analysis | first_pass | second_pass | order | units |+--------+------+----------+------------+-------------+-------+-------+| apple  |    1 | full     | 12.1       | 20.1        |     2 | g     || apple  |    2 | full     | 7.1        | 12.0        |     1 | g     || apple  |    1 | partial  | 14.3       | 13.1        |     3 | g     || orange |    1 | full     | 19.1       | 20.1        |     2 | g     || orange |    2 | full     | 17.1       | 18.5        |     1 | g     || orange |    1 | partial  | 23.4       | 22.7        |     3 | g     || grape  |    1 | full     | 23.1       | 14.1        |     3 | g     || grape  |    2 | full     | 17.2       | 17.1        |     2 | g     || grape  |    1 | partial  | 19.1       | 19.4        |     1 | g     |+--------+------+----------+------------+-------------+-------+-------+

我想添加几列:

• 一个 bool 列，用于指示该测试和分析的 second_pass 值是否是所有水果类型中最高的。
• 另一列列出了每种测试和分析组合中哪些水果具有最高的 second_pass 值。

使用这个逻辑，我想得到下表:

+--------+------+----------+------------+-------------+-------+-------+---------+---------------------+| fruit  | test | analysis | first_pass | second_pass | order | units | highest |   highest_fruits    |+--------+------+----------+------------+-------------+-------+-------+---------+---------------------+| apple  |    1 | full     | 12.1       | 20.1        |     2 | g     | true    | ["apple", "orange"] || apple  |    2 | full     | 7.1        | 12.0        |     1 | g     | false   | ["orange"]          || apple  |    1 | partial  | 14.3       | 13.1        |     3 | g     | false   | ["orange"]          || orange |    1 | full     | 19.1       | 20.1        |     2 | g     | true    | ["apple", "orange"] || orange |    2 | full     | 17.1       | 18.5        |     1 | g     | true    | ["orange"]          || orange |    1 | partial  | 23.4       | 22.7        |     3 | g     | true    | ["orange"]          || grape  |    1 | full     | 23.1       | 22.1        |     3 | g     | false   | ["orange"]          || grape  |    2 | full     | 17.2       | 17.1        |     2 | g     | false   | ["orange"]          || grape  |    1 | partial  | 19.1       | 19.4        |     1 | g     | false   | ["orange"]          |+--------+------+----------+------------+-------------+-------+-------+---------+---------------------+

我是 pandas 的新手，所以我确定我错过了一些非常简单的东西。

Answer 1

您可以返回 boolean 值，其中 second_pass 等于 group max，如 idxmax 只返回第一次出现的 max:

df['highest'] = df.groupby(['test', 'analysis'])['second_pass'].transform(lambda x: x == np.amax(x)).astype(bool)

然后使用 np.where 来捕获所有具有 group max 的 fruit 值，并且 merge 结果到你的 DataFrame 中，像这样:

highest_fruits = df.groupby(['test', 'analysis']).apply(lambda x: [f for f in np.where(x.second_pass == np.amax(x.second_pass), x.fruit.tolist(), '').tolist() if f!='']).reset_index()
df =df.merge(highest_fruits, on=['test', 'analysis'], how='left').rename(columns={0: 'highest_fruit'})

最后，为了您的跟进:

first_pass = df.groupby(['test', 'analysis']).apply(lambda x: {fruit: x.loc[x.fruit==fruit, 'first_pass'] for fruit in x.highest_fruit.iloc[0]}).reset_index()
df =df.merge(first_pass, on=['test', 'analysis'], how='left').rename(columns={0: 'first_pass_highest_fruit'})

得到:

  analysis  first_pass   fruit  order  second_pass  test units highest  \
0     full        12.1   apple      2         20.1     1     g    True   
1     full         7.1   apple      1         12.0     2     g   False   
2  partial        14.3   apple      3         13.1     1     g   False   
3     full        19.1  orange      2         20.1     1     g    True   
4     full        17.1  orange      1         18.5     2     g    True   
5  partial        23.4  orange      3         22.7     1     g    True   
6     full        23.1   grape      3         14.1     1     g   False   
7     full        17.2   grape      2         17.1     2     g   False   
8  partial        19.1   grape      1         19.4     1     g   False   

     highest_fruit             first_pass_highest_fruit  
0  [apple, orange]  {'orange': [19.1], 'apple': [12.1]}  
1         [orange]                   {'orange': [17.1]}  
2         [orange]                   {'orange': [23.4]}  
3  [apple, orange]  {'orange': [19.1], 'apple': [12.1]}  
4         [orange]                   {'orange': [17.1]}  
5         [orange]                   {'orange': [23.4]}  
6  [apple, orange]  {'orange': [19.1], 'apple': [12.1]}  
7         [orange]                   {'orange': [17.1]}  
8         [orange]                   {'orange': [23.4]}